+0

Triangle Area

+5
1209
6

Which has a larger area

A triangle with dimensions 29 cm by 29 cm by 40 cm, or a triangle with dimensions 29 cm by 29 cm  by 42 cm

Pls explain solution

Thanks

Feb 12, 2016

#3
+111321
+10

They have exactly the same area......believe it, or not.....!!!!!

Using Heron's formula to prove this, we have, for the 29,29,40 triangle

sqrt [49(20)(20)(9) ] =  about 420 cm^2

And for the 29,29, 42 triangle, we have

sqrt [ (50(21)(21)(8) ]  =  420 cm^2

Here is the explanation for the derivation of Heron's Formula along with a calculator to verify the above results :

http://www.mathsisfun.com/geometry/herons-formula.html

Feb 12, 2016
edited by CPhill  Feb 12, 2016
edited by CPhill  Feb 12, 2016

#1
0

These are isosceles triangles. Area=1/2 X base X height

So, one with base 42 would have a larger area.

Feb 12, 2016
#2
+5

.....but the HEIGHT changes with the bigger BASE resulting in BOTH triangles having the same area.

Feb 12, 2016
#3
+111321
+10

They have exactly the same area......believe it, or not.....!!!!!

Using Heron's formula to prove this, we have, for the 29,29,40 triangle

sqrt [49(20)(20)(9) ] =  about 420 cm^2

And for the 29,29, 42 triangle, we have

sqrt [ (50(21)(21)(8) ]  =  420 cm^2

Here is the explanation for the derivation of Heron's Formula along with a calculator to verify the above results :

http://www.mathsisfun.com/geometry/herons-formula.html

CPhill Feb 12, 2016
edited by CPhill  Feb 12, 2016
edited by CPhill  Feb 12, 2016
#4
+111321
+10

Here's the visual proof of this.....

Draw a circle centered at the origin with a radius of 29  with the specified triangles inscribed in the bottom half of this circle

Triangle ECD  is isosceles with sides = 29, a base of 42  and a height, GC = 20

Triangle  BCA  is iscosceles with sides = 29, a base of 40  and a height,  FC = 21

So....area of ECD = [42 * 20]/2  =  840/2 = 420 cm^2

And BCA has an area of [ 40 * 21] / 2  = 840/2  =  420 cm^2

Thus   area of ECD  = area of BCA   ....!!!!

Feb 12, 2016
edited by CPhill  Feb 12, 2016
#5
+109518
0

Thanks Chris,

That is a very unexpected outcome :))

Feb 14, 2016
#6
+111321
0

One more comment about this problem....we will always have integer solutions and equal triangles when the quantties involved form Pythagorean Triples.......specifically......

If the isoceles sides of the triangles  =   "c"  in the formula  a^2 + b^2  = c^2

And if we let the base of one  triangle  = two times the length of one of the legs in the triple.....then the height will be equal to the remaining number in the triple.

Feb 14, 2016